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zubka84 [21]
4 years ago
15

A drug contains 92 marvels of which 20 are blue 8 are red and the rest are green.what is the ratio of blue marbles to green marb

les
Mathematics
1 answer:
Murljashka [212]4 years ago
7 0

Answer:

20:64

Step-by-step explanation:

total = 92

red = 8

blue = 20

green = x

20:X

------------

92 - 28 (blue + red) will give you the amount of green, which is 64

so the blue:green ratio is 20:64

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Simplify x3 / x <br> A)x<br> B)1/x2<br> C)x2<br> D)1/x
anygoal [31]
The answer is x^2, which is C), because x^2 * x = x^3
6 0
3 years ago
A tank contains 1000L of pure water. Brine that contains 0.02kg of salt per liter enters the tank at a rate of 5L/min. Also, bri
tankabanditka [31]

Answer:

51.877 kg

Step-by-step explanation:

a

See attachment for calculations

b

The amount of salt in the tank after 4 hours is gotten by simply replacing t in the equation from a, with 4 hours. Since the t gotten in a above is in minutes, we'd convert hours to minute as well. If we do that, we have

t = 4 hours = 4 * 60 minutes = 240 minutes

y = 160/3 (1 - e^((-3 * 240)/200))

y = 160/3 (1 - e^(-720/200))

y = 160/3 (1 - 0.0273)

y = 160/3 (0.9727)

y = 51.877 kg

Therefore, the amount of salt in the tank after 4 hours is 51.877 kg

3 0
3 years ago
Find the area of the trapezoid.
UkoKoshka [18]
The answer is 68cm^2. Hope this helps!
4 0
3 years ago
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab
Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, H_0 : \mu = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, H_1 : \mu < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

       T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = \frac{738.44-750}{\frac{38.2}{\sqrt{50} } } ~ t_4_9

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0
3 years ago
X and y intercepts g(x)=9x-10
anygoal [31]

the y intercept is -10

the x intercept is

0=9x-10

10=9x

x=10/9

6 0
3 years ago
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